Physics: Calculate x(t), T & Kinetic Energy, Find Force & Direction of Block

[nyyfan0729]

A block of mass 0.5kg moving on a horizontal frictionless surface at 2.0 m/s collides with and sticks to a massless pan attached to the end of a horizontal ideal spring whose spring constan is 32 N/m.
a) Determine the function for x(t), the displacement from equilibrium position as a function of time.
b) What is the period T, of the subsequent oscillations?
c) What is the kinetic energy of the mass 4.0 sec after it collides with the spring?
d) What force is exerted by the spring on the block at t=1.2 sec? Which way is the block moving? Explain

 

[dicerandom]

Well, what have you done? Which parts are you having problems with?

 

[kyle8921]

a) To determine the function for x(t), we can use the equation for simple harmonic motion, x(t) = A*cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase constant. In this case, the block is initially at rest at the equilibrium position, so φ = 0. The amplitude can be calculated using the conservation of energy equation, where the potential energy stored in the spring is equal to the initial kinetic energy of the block. Thus, A = √(2*m*k*x_0), where m is the mass of the block, k is the spring constant, and x_0 is the initial displacement of the block. The angular frequency can be calculated using the equation ω = √(k/m). Putting these values into the equation for x(t), we get x(t) = √(2*m*k*x_0)*cos(√(k/m)*t).

b) The period T of the subsequent oscillations can be calculated using the equation T = 2π/ω. In this case, T = 2π/√(k/m).

c) The kinetic energy of the mass 4.0 sec after it collides with the spring can be calculated using the equation KE = 0.5*m*v^2, where m is the mass and v is the velocity. In this case, the mass is still 0.5 kg and the velocity is given as 2.0 m/s, so the kinetic energy is KE = 0.5*0.5*2.0^2 = 0.5 J.

d) To calculate the force exerted by the spring on the block at t=1.2 sec, we can use the equation F = -k*x, where k is the spring constant and x is the displacement from equilibrium position. At t=1.2 sec, the block is at a displacement of x = A*cos(√(k/m)*1.2) = A*cos(1.2*√(k/m)). Using the values for A and ω calculated in part a), we can calculate the force as F = -(32 N/m)*(√(2*0.5*32*0.5)*cos(1.2*√(32/0.5))) = -16*cos(1.2*4